Find the equation of the line that is perpendicular to Y=6x+3 and passes through the point (-24,5)
1 answer:
The slope of the given line is the coefficient of x, 6. The slope of the perpendicular line will be the negative reciprocal of that, -1/6. Then the point-slope form of the equation of the line is
y = m(x -h) +k . . . . . . line with slope m through point (h, k)
You want m=-1/6, (h, k) = (-24, 5), so your equation is
y = (-1/6)(x +24) +5
y = (-1/6)x +1
y = m(x -h) +k . . . . . . line with slope m through point (h, k)
You want m=-1/6, (h, k) = (-24, 5), so your equation is
y = (-1/6)(x +24) +5
y = (-1/6)x +1
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Answer:
Step-by-step explanation:
X^2-6x+5=x^2-5x-x+5=x(x-5)-(x-5)=(x-5)(x-1)
Then we have (x-5)(x-1)=0 if x=5 or x=1.
Intersections on x are points (1,0) and (5,0), middle is (3,0).
Intersection with y is when put x=0 in equation, so you will get y=0-6*0+5, y=5. The point is (0,5).
From picture symmetry is line x=3.
